8. Sodium thiosulfate, Na2S2O3, is an important reagent for...

8. Sodium thiosulfate, Na2S2O3, is an important reagent for titrations. Its solutions can be standardized by titrating the iodine released when a weighed amount of potassium hydrogen iodate, KH(IO3)2 (389.912 g/mol), is allowed to react with excess potassium iodide in acidic solutions. The net ionic equations are: production of iodine from KH(IO3)2: IO3? + 5 I? + 6 H? ? 3 I2 + 3 H2O titration of iodine: I2 + 2 S2O3²? ? 2 I? + S4O6²? What is the molarity of a sodium thiosulfate solution if 29.65 mL are required to titrate the iodine released from 0.1583 g of KH(IO3)2?

Transcript

00:01 Hello students welcome here in this question we have a reaction or titration of iodine with sodium thiosulfate.

00:12 So titration agnest iodine agnest sodium thiosulfate.

00:18 Okay so the iodine that is coming from k -potassium iodate is so the amount of potassium iodate is.

00:32 Iodate is given 0 .1 to 78 grams.

00:37 So the iodine that is produced from this sodium potassium iodate come to completely titrated amount of this sodium iodide sodium thiosulfide required molarity concentration we have to find out for this one.

00:57 So volume is given 39 so here the volume is 32.

01:02 2 .96 ml is given.

01:06 We have to find out the molarity of sodium thiosulfate that is required to iodide, completely iodide, the completely titrated, the iodine that is produced from potassium iodide.

01:23 So the reaction we will see first, iodide minus plus 5 moles of iodide ion.

01:34 6h plus gives to s2 3 i2 is produced here and 3 moles of h2 will produce it and the titration reaction is i 2 with 2 moles of h2 o3 minus 2 gives to h2 2 i minus plus s 4 o 6 minus 2 so these 2 are the reactions that are happening while the production of iodine from potassium iodate and the titration of iodine -agnased sodium thiosulfate.

02:12 So you will see the first number of moles of this potassium iodate we will see.

02:18 So one mole of potassium iodide, i -o -3 minus, gives two moles of iodide ions, right? so here the number of moles we will find out.

02:35 0.

02:36 Here we have 0 .1278 grams.

02:42 So number of moles is equal to the given mass divided by molar mass.

02:48 Okay.

02:48 So molar mass is 389 .9 .9 .2 is given.

02:54 So that is equal to 3 .277 multiplied by 10 minus 4 moles produced.

03:01 So 3 .2 .2 .27 multiplied by 10 power minus 4 moles, how much of iodate will produce here? so number of moles of iodate ions produced is equal to number of moles of iodate ions is equal to 2 multiplied by 3 .277 multiplied by 10 power minus 4 moles...

Question

Please give Ace some feedback